SVM - Understanding the math - Part 2

This is Part 2 of my series of tutorial about the math behind Support Vector Machines.
If you did not read the previous article, you might want to start the serie at the beginning by reading this article: an overview of Support Vector Machine.

In the first part, we saw what is the aim of the SVM. Its goal is to find the hyperplane which maximizes the margin.

But how do we calculate this margin?

SVM = Support VECTOR Machine

In Support Vector Machine, there is the word vector.That means it is important to understand vector well and how to use them.

Here a short sum-up of what we will see today:

What is a vector?

its norm

its direction

How to add and subtract vectors ?

What is the dot product ?

How to project a vector onto another ?

Once we have all these tools in our toolbox, we will then see:

What is the equation of the hyperplane?

How to compute the margin?

What is a vector?

If we define a point in we can plot it like this.

Figure 1: a point

Definition: Any point , in specifies a vector in the plane, namely the vector starting at the origin and ending at x.

This definition means that there exists a vector between the origin and A.

Figure 2 - a vector

If we say that the point at the origin is the point then the vector above is the vector . We could also give it an arbitrary name such as .

Note: You can notice that we write vector either with an arrow on top of them, or in bold, in the rest of this text I will use the arrow when there is two letters like and the bold notation otherwise.

Ok so now we know that there is a vector, but we still don't know what IS a vector.

Definition: A vector is an object that has both a magnitude and a direction.

We will now look at these two concepts.

1) The magnitude

The magnitude or length of a vector is written and is called its norm.For our vector , is the length of the segment

Understanding the definition

Naive definition 1: The direction of the vector is defined by the angle with respect to the horizontal axis, and with the angle with respect to the vertical axis.

This is tedious. Instead of that we will use the cosine of the angles.

In a right triangle, the cosine of an angle is defined by :

In Figure 4 we can see that we can form two right triangles, and in both case the adjacent side will be on one of the axis. Which means that the definition of the cosine implicitly contains the axis related to an angle. We can rephrase our naïve definition to :

Naive definition 2: The direction of the vector is defined by the cosine of the angle and the cosine of the angle .

Now if we look at their values :

Hence the original definition of the vector . That's why its coordinates are also called direction cosine.

Computing the direction vector

We will now compute the direction of the vector from Figure 4.:

and

The direction of is the vector

If we draw this vector we get Figure 5:

Figure 5: the direction of u

We can see that as indeed the same look as except it is smaller. Something interesting about direction vectors like is that their norm is equal to 1. That's why we often call them unit vectors.

The sum of two vectors

Figure 6: two vectors u and v

Given two vectors and then :

Which means that adding two vectors gives us a third vector whose coordinate are the sum of the coordinates of the original vectors.

You can convince yourself with the example below:

Figure 7: the sum of two vectors

The difference between two vectors

The difference works the same way :

Figure 8: the difference of two vectors

Since the subtraction is not commutative, we can also consider the other case:

Figure 9: the difference v-u

The last two pictures describe the "true" vectors generated by the difference of and .

However, since a vector has a magnitude and a direction, we often consider that parallel translate of a given vector (vectors with the same magnitude and direction but with a different origin) are the same vector, just drawn in a different place in space.

So don't be surprised if you meet the following :

Figure 10: another way to view the difference v-u

and

Figure 11: another way to view the difference u-v

If you do the math, it looks wrong, because the end of the vector is not in the right point, but it is a convenient way of thinking about vectors which you'll encounter often.

A few words on notation

The dot product is called like that because we write a dot between the two vectors.
Talking about the dot product is the same as talking about

the inner product (in linear algebra)

scalar product because we take the product of two vectors and it returns a scalar (a real number)

The orthogonal projection of a vector

Given two vectors and , we would like to find the orthogonal projection of onto .

Figure 16

To do this we project the vector onto

Figure 17

This give us the vector

Figure 18 : z is the projection of x onto y

By definition :

We saw in the section about the dot product that

So we replace in our equation:

If we define the vector as the direction of then

and

We now have a simple way to compute the norm of the vector .
Since this vector is in the same direction as it has the direction

And we can say :

The vector is the orthogonal projection of onto .

Why are we interested by the orthogonal projection ? Well in our example, it allows us to compute the distance between and the line which goes through .

Figure 19

We see that this distance is

The SVM hyperplane

Understanding the equation of the hyperplane

You probably learnt that an equation of a line is : . However when reading about hyperplane, you will often find that the equation of an hyperplane is defined by :

How does these two forms relate ?
In the hyperplane equation you can see that the name of the variables are in bold. Which means that they are vectors ! Moreover, is how we compute the inner product of two vectors, and if you recall, the inner product is just another name for the dot product !

Note that

is the same thing as

Given two vectors and

The two equations are just different ways of expressing the same thing.

It is interesting to note that is , which means that this value determines the intersection of the line with the vertical axis.

Why do we use the hyperplane equation instead of ?

For two reasons:

it is easier to work in more than two dimensions with this notation,

the vector will always be normal to the hyperplane(Note: I received a lot of questions about the last remark. will always be normal because we use this vector to define the hyperplane, so by definition it will be normal. As you can see this page, when we define a hyperplane, we suppose that we have a vector that is orthogonal to the hyperplane)

And this last property will come in handy to compute the distance from a point to the hyperplane.

Compute the distance from a point to the hyperplane

In Figure 20 we have an hyperplane, which separates two group of data.

Figure 20

To simplify this example, we have set .

As you can see on the Figure 20, the equation of the hyperplane is :

which is equivalent to

with and

Note that the vector is shown on the Figure 20. (w is not a data point)

We would like to compute the distance between the point and the hyperplane.

This is the distance between and its projection onto the hyperplane

Figure 21

We can view the point as a vector from the origin to .
If we project it onto the normal vector

Figure 22 : projection of a onto w

We get the vector

Figure 23: p is the projection of a onto w

Our goal is to find the distance between the point and the hyperplane.
We can see in Figure 23 that this distance is the same thing as .
Let's compute this value.

We start with two vectors, which is normal to the hyperplane, and which is the vector between the origin and .

Let the vector be the direction of

is the orthogonal projection of onto so :

Compute the margin of the hyperplane

Now that we have the distance between and the hyperplane, the margin is defined by :

We did it ! We computed the margin of the hyperplane !

Conclusion

This ends the Part 2 of this tutorial about the math behind SVM.
There was a lot more of math, but I hope you have been able to follow the article without problem.

I am passionate about machine learning and Support Vector Machine. I like to explain things simply to share my knowledge with people from around the world. If you wish you can add me to linkedin, I like to connect with my readers.

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122 thoughts on “SVM - Understanding the math - Part 2”

Are you planning to tell about support vectors, non-linear kernels and optimization (I mean finding the minimum of the distance from the hyperplane to the suport vectors) in this tutorial? It seems that one need to know optimization theory in depth to understand this algorithm. It would be nice to see the simple explanation of what the algorithm is doing actually.

Yes that is what I am planning to do. However optimization theory is indeed very important to understand the algorithm and I am still figuring out how to explain SVM without going too deep into details.

Hello Alexandre.. Great explanation.. Thanks..
I have a question on figure19.

My interpretation below:
We are assuming ∥x−z∥ to be the hypotenuse but actually it is not the case right? the vector opposite to ∥x−z∥ is hypotenuse (X). So the formula following that should substract the distance from the hypotenuse instead of adding.

When I do the calculation in figure, I say, let us compute the norm of the vector x-z. Let us call this vector k. What are the coordinates of k? k(5-1, 3-4) which gives us k(4,-1), then we compute the norm as usual ∥k∥ = sqrt(4^2+1^2) = sqrt(17)

I was trying to understand SVM from a very long time. your blog really helped me a lot and now I know what I am dealing with. your tutorial not only helped in understanding the mathematical jargon but also give me the clear perspective of what I am doing.
Thanks a lot!!

Thanks for the comment Shyam. I am afraid that recently I have spent most of my time on kaggle competitions and playing with convolutionnal neural networks. I will try to write the following part in the coming weeks in order to no achieve this work.

This is the best blog about SVM I have seen ever, help me so much, thank you very much, look forward to excellent part 3. BTW, "To simplify this example, we have set ", do you mean that setting the start point of vector at origin?

Thanks for your kind comment. No this does not mean setting the start point of the vector at the origin. We could place it somewhere else because we often consider that the parallel translate of a given vector is the same vector (this is illustrated in the section about the difference of two vectors)In the definition of the equation of a hyperplane the vector is a 3-dimensional vector . By setting to 0 we can do the remaining calculations with a 2-dimensional vector. Because the definition says that and we use instead, it removes the intercept term from the equation. As a result the hyperplane passes through the origin. In the Part 3 I wrote in more details about the hyperplane equation, things should be easier to understand.

Thank you. You find these two vectors by continuing the reasoning.
We want to express the equation y-ax−b=0 with a dot product between two vectors.
The dot product is the sum of several products. In our case there is two minus signs, so there is three elements being summed together. Our vectors will have three elements each. Then we transform the equation to display these products: y−ax−b=0 is equivalent to y*1−a*x−b*1=0 and then we transform the differences into sums : y*1+(−a)*x+(−b)*1=0

Thank you for very explicit explanation.
I have a question since I have no background knowledge about cosine, sine, etc.
Shouldn't Direction of vector be just angle of the triangle? I am just curious what cos(β)= adjacent/hypotenuse formula fundamentally means?

No because you can have another vector with the same angle between the axis and itself but with the vectors pointing in another direction. By using cosine, we use the length of the adjacent and hypotenuse and as we are using coordinates we obtain a vector pointing in the same direction.

Thank for you blog, that is great. However, i have a question about the W(-b,-a,1) and X(1,x,y),
the transpose of W is a column vector and X is a row vector, the result of [ column * row] is a matrix that size is (3,3), can you tell me where i missing?

Very Nice and crystal clear explaination i have ever found on internet.
It will be very helpful if you give some practical demostration of how SVM and other
learning algorithms can be implemented and interpreted on various platforms like weka and orange.What is confusion matrix and ROI. How that Wt-b equation is generated etc.
Giving practical demonstration will be very helpful.

Phenomenal explanation of SVMs. Thanks a lot for taking the time to write and publish this. I was wondering if you would mind if I used brief excerpts of your content. I am preparing a few slides for a course that I will be teaching and found some of your images and explanations very helpful. I'll be sure to include citations and a reference to your blog posts.

Speechless, this is downright simple to understand. This makes SVM move from very hard to simply understandably, thanks a lot mate. At least now i have an idea of what's happening behind the scenes of svm.SVC().fit(),
Great work.

You are amazing. Thanks a lot for this.. Because of lack of enough maths background i have having difficulty reaching here . You helped a lot. Is it possible for you to explain the justification of langrange's multipliers as well as further explanation of SVM.

In your example: x2 = 1/3 * x1 + 1, is in the form: x2 = a * x1 + b. To get the normal vector you just get the vector w(a,-1). So in this case, we define w(1/3,-1), x(x1,x2) and b = 1. And you can see that wx+b=0 is equivalent to x2 = 1/3 * x1 + 1.

Now if we plot, the vector w(1/3,-1), we can start to draw it where we want. I could start drawing at the origin x(0,0) or I can start drawing it directly on the hyperplane. I choose to do so, and I start drawing it at x(1,1+1/3).
As you can see in the figure: it is normal to the hyperplane.

Where I chose to start drawing it, does not change the fact that it is normal to the hyperplane.

Given an hyperplane having the equation wx+b=0 with vectors w(w0,w1) and x(x0,x1). b is the distance between the vertical axis and the origin only when the value w1 of the weight vector is equal -1. Indeed, when we transform this hyperplane equation to a line equation of the form y=ax+c we get a = -w0/w1 and c = -b/w1. Some books represent b as being the distance between the origin and the hyperplane, but I think this is true only under certain conditions, at least that is what I found when trying to verify it by myself using the first formula of this article using formulas from this page.

Thank you a thousand times............You explained Lagrange multipliers in the best way in the world.....
can you introduce me some useful books which I can read and get more information about classification?

Awesome tutorial, TAL man!
Just a small suggestion, when you give a link like for "cumulative", "dot product" , can you change a code of your site such that after clicking on this link, it gets open in different tab instead of opening in current tab.

x+.w+b= +1
x_.w+b = -1
why it is always equal to +1 and -1 for positive and negative support vector respectively.?. How to normalize this distance of hyperplane to support vectors?. Once we normalize, it always remains same for any kind of data. could you explain?. I am not clear about the distance between hyperplane and support vectors

It is always equal to +1 and -1 because we are free to select w for which it will be the case (we can rescale w and b and keep the same hyperplane). So we decide arbitrarily to select among the ones for which it is equal to +1 and -1 because it will make the following computation easier.

I had taken machine learning as my final semester research topic.For about a month I was unable to decide which topic to specifically decide to work upon.After a month my guide told me to work upon SVM in image processing.I had a little knowledge about SVM but the math part was very difficult.It was intricate as to say.
Finally I came upon this blog and found it very help , the math was really very detailed yet simple to understand.Thanks for such a nice article the 6 part series has helped me a lot in understanding machine learning as a whole.

As stated in the comments before, this is by definition. We defined the hyperplane with this equation, so w is normal. If you read the example in this page: you can see that they say "Let’s also suppose that we have a vector that is orthogonal (perpendicular) to the plane".

We can see that ww as indeed the same look as uu except it is smaller. Something interesting about direction vectors like ww is that their norm is equal to 1. That's why we often call them unit vectors. -- why direction vectors like 'w' is that their norm is equal to 1? can you please explain?

I understood the proceeding for calculate the distance but there is a method more simple for do it: distance= dot(A,w)/||w|| Is there a particular reason for prefer your described method rather than my simpler?

Hi, great tutorial I am learning lots from this. I have a doubt though, to find the orth proj of a on X1=-2x2 you have used w(1,2), how did u arrive to this value of w? The ortho project of a onto hyper plane X1=-2x2 should be same from any point normal to it, but your equations are dependent on the values of w and if I were to choose a different line normal to hyperplane I would get a diff magnitude won't I ? It would be great if you can explain

Yes indeed but it does not matter if you get a different magnitude because you are trying to minimize the distance and you would use the same w in all your calculations. This ability to choose the w arbitrarily is used later when we define the optimization problem. I explain this in detail in my ebook.

Start from the hyperplane and follow the dotted line until you reach A. This is the distance we want to find. Start from the hyperplane (0,0) and follow the line until you reach the end of p. You can see that this distance is the same. You have to see that p is the vector starting at (0,0 and finishing at (4,2). ||p|| the length of p, so ||p|| is this distance we are looking for

All the mathematics and calculus would be more clear if you can show it with some working example.
Example: We have a data set that has 2 explanatory variables i.e x1 and x2 and one binary outcome variable y. Then how would we calculate hyper plane and w values.

Thank you for this great tutorial!
I have recently started my hopeless journey to understanding SVM. I'm wondering why this example starts hyperplane calculations from point origin (0,0)? how do equations change in situation when every data point is of the right side of (0,0). Where do we get the starting point?

hii...first of all i would like to thank you for sharing a wonderful site like this. I found this article quite interesting and useful for me.But, unfortunately I would like you bring your kind attention that, in this article, it is showing "MATH PROCESSING ERROR" instead of showing the actual math equation or any numeric.

Hope you will consider this situation and would make the necessary corrections and would reply me.

Hello, Thank you for your message. The equations are displayed correctly on my computer. Maybe you have a problem with your web browser. I am using Chrome and everything looks good. You could try with another browser.

I am sorry but I am still confused about vector. I understand the math and the numbers you have shown here, but I cannot visualise vector yet in real world. When you add 2 and 3 apples I can visualise 5 apples. When you add two vectors, subtract them, dot, cross product, I understand the math, but I still cant visualise it.

Well, in figure 7 you can visualize that the sum of the two vectors (the two arrows) produce a third vector (the middle arrow). Here the arrow, is a representation of the vector. The same way I could represent the number one with the picture of one apple, and the number five with the picture of five apples. What is important to understand is that both a number and a vector should be considered as mathematical objects. And that these object have some properties. I can recommend you to read the books Mathematics a very short introduction and How to study as a mathematics major to get a better understanding of how you can consider mathematical objects.